Contoh Soal 5 Fungsi Kuadrat (Kelas X/Fase E Semester Genap) Tahun 2024

$\begin{array}{l}\\ 21.& \text{Jika}y=\sqrt{x^2-2x+1}+\sqrt{x^2+2x+1}\\ & \text{maka nilai}y=....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle {\displaystyle 2x}}& & & & & \text{D}.& {\displaystyle |x-1|+|x+1|}\\ \text{B}.& {\displaystyle 2(x+1)}& & \text{C}.& {\displaystyle 0}& & \text{E}.& {\displaystyle \text{tidak ada yang benar}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{D}}\\ & \text{Diketahui}y=\sqrt{x^2-2x+1}+\sqrt{x^2+2x+1}\\ & \iff y=\sqrt{(x-1{)}^2}+\sqrt{(x+1{)}^2}\\ & \iff y=|x-1|+|x+1|\end{array}$.

$\begin{array}{l}\\ 22.& \text{Jika}x\text{bilangan real dan}4y^2+4xy+x+6=0,\\ & \text{maka nilai semua nilai}x\text{agar nilai}y\text{real adalah}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle x\le -2\text{atau}x\ge 3}& & & & & \text{D}.& {\displaystyle -3\le x\le 2}\\ \text{B}.& {\displaystyle x\le 2\text{atau}x\ge 3}& & & {\displaystyle }& & \text{E}.& {\displaystyle -2\le x\le 3}\\ \text{C}.& {\displaystyle x\le -3\text{atau}x\ge 2}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{A}}\\ & \begin{array}{rl}& 4y^2+(4x)y+(x+6)=0\\ & \text{agar}y\text{real, maka}D=b^2-4ac\ge 0\\ & (4x{)}^2-4(4)(x+6)\ge 0\\ & \iff 16x^2-16(x+6)\ge 0\\ & \iff x^2-x-6\ge 0\\ & \iff (x-3)(x+2)\ge 0\\ & \iff {\displaystyle x\le -2\text{atau}x\ge 3}\end{array}\\ \\ & \text{Jika kita diletakkan dalam garis bilangan}\\ & \text{akan tampak seperti berikut}\\ \\ & \begin{array}{rl}& \underset{\begin{array}{c}\Downarrow \\ \Downarrow \end{array}}{\text{Daerah Positif}}\\ & \begin{array}{c}\\ & & & & & & \\ ++& ++& --& --& --& ++& ++\\ & -{\displaystyle 2}& & & & 3& & \mathbf{\text{Sumbu-X}}\end{array}\\ & \begin{array}{c}\text{dengan}\\ \text{titik}\\ \text{uji}\\ x=0\end{array}\stackrel{\begin{array}{c}\Uparrow \\ \Uparrow \end{array}}{\text{Daerah negatif}}\stackrel{\begin{array}{c}\Uparrow \\ \Uparrow \end{array}}{\text{Daerah positif}}\end{array}\end{array}$.

$\begin{array}{l}\\ 23.& \text{Diketahui}f(x)={\displaystyle \frac{x(x-1)}{2},\text{nilai}f(x+2)=....}\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle f(x)+f(2)}& & & & & \text{D}.& {\displaystyle \frac{xf(x)}{x+2}}\\ \text{B}.& {\displaystyle (x+2)f(x)}& & & {\displaystyle }& & \text{E}.& {\displaystyle {\displaystyle \frac{(x+2)f(x+1)}{x}}}\\ \text{C}.& {\displaystyle x(x+2)f(x)}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{E}}\\ & f(x)={\displaystyle \frac{x(x-1)}{2},}\\ & f(x+1)={\displaystyle \frac{(x+1)(x)}{2}\iff {\displaystyle \frac{x+1}{2}=\frac{f(x+1)}{x},\text{maka}}}\\ & \begin{array}{rl}f(x+2)& ={\displaystyle \frac{(x+2)((x+2)-1)}{2}=\frac{(x+2)(x+1)}{2}}\\ & ={\displaystyle \frac{f(x+1)}{x}.(x+2)={\displaystyle {\displaystyle \frac{(x+2)f(x+1)}{x}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 24.& \text{Jika}{r}_1\text{dan}{r}_2\text{merupakan dua penyelesaian }\\ & \text{dari persamaan}{x}^2+px+8=0,\text{maka}....\\ & \begin{array}{llllll}\text{A}.& {\displaystyle |r_1+r_2|>4\sqrt{2}}& & & \text{D}.& {\displaystyle r_1<0\text{dan}{r}_2<0}\\ \text{B}.& {\displaystyle \left|r_1\right|>3\text{dan}\left|r_2\right|>3}& {\displaystyle }& & \text{E}.& {\displaystyle -|r_1+r_2|<4\sqrt{2}}\\ \text{C}.& {\displaystyle \left|r_1\right|>2\text{dan}\left|r_2\right|>2}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{A}}\\ & \begin{array}{rl}& \text{Diketahui PK}:x^2+px+8=0\\ & \text{dengan}\{\begin{array}{c}{r}_1\\ r_2\end{array}\text{akar-akarnya, maka}\\ & \bullet r_1+r_2=-p\\ & \bullet r_1\times r_2=8\\ & \text{Asumsikan kedua akarnya real dan berbeda,}\\ & \text{sehingga kita pilih nilai}D=b^2-4ac>0\\ & \iff (-p{)}^2-4.1.8>0\\ & \iff p^2-32>0\\ & \iff (|r_1+r_2|+\sqrt{32})(|r_1+r_2|-\sqrt{32})>0\\ & \iff |r_1+r_2|<-\sqrt{32}\text{atau}|r_1+r_2|>\sqrt{32}\\ & \stackrel{\begin{array}{c}\Downarrow \\ \Downarrow \end{array}}{\mathbf{\text{tidak mungkin}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 25.& \text{Jika titik}(1,y_1)\text{dan}(-1,y_2)\text{terletak pada}\\ & \text{grafik}y=a{x}^2+bx+c\text{dan}{y}_1-y_2=-6\\ & \text{maka nilai}b=....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle {\displaystyle -3}}& & & & & \text{D}.& {\displaystyle \sqrt{ac}}\\ \text{B}.& {\displaystyle 0}& & \text{C}.& {\displaystyle 3}& & \text{E}.& {\displaystyle {\displaystyle \frac{a+c}{2}}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{D}}\\ & \text{Diketahui}y=a{x}^2+bx+c,\text{untuk}\\ & \bullet (1,y_1)\Rightarrow y_1=a+b+c\\ & \bullet (-1,y_1)\Rightarrow y_2=a-b+c\\ & \mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}-\\ & y_1-y_2=2b=-6\Rightarrow b=-3\end{array}$.